Help with Inserting Data in table

tigomark

This should be something small and easy to fix but I have been staring at this for too long. I am trying to insert data from a form to a video rental database. The info is being held in a temp dbase until the user wants to rent. All values show if I call them to the next page but I get a die error when I try to insert into the dbase.

Here's the code:

$db_name = "$list";

$table_name = "$tempdbase";

$connection = @mysql_connect("$host", "$username", "$passwrd") or
die("Couldn't connect.");
//works so far

$db = @mysql_select_db($db_name, $connection) or die("Couldn't select database.");
//still good

$sql = "
INSERT INTO $table_name
(id, Title, Genre, Decade, Director, Actor1First, Actor1Last,
Actor2First, Actor2Last, Actor3First, Actor3Last, NewRelease,
Description, ImageName, EmployeeName, OwnersEmail, Rented, RentedEmail, By)
VALUES
(\"id\", \"$Title\", \"$Genre\", \"$Decade\", \"$Director\",
\"$Actor1First\", \"$Actor1Last\", \"$Actor2First\", \"$Actor2Last\",
\"$Actor3First\", \"$Actor3Last\", \"$NewRelease\", \"$Description\",
\"$ImageName\", \"$EmployeeName\", \"$OwnersEmail\", \"$Rented\", \"$Email\",
\"$FName $LName\")

$result = @($sql, $connection) or die ("Not executing query");
//This is where I die.

again I think it is something small, but I must be missing it.

Thank you all for your help. It was a syntax error with the varible $By. Sql did not like me using one of it's opperators as a varible. I knew it had to be something small. Thank you for your help!!!

aksival

Temporarily comment-out the 'or die...' part and echo out the resulting mysql_error()...this will give you some insight as to what's actually happening with the query.

tigomark

I have done that and the script is not inserting into the dbase at all. It seems to connect and select the dbase but it will not perform the query.

aksival

What is your output when you change

$result = @mysql_query($sql, $connection) or die ("Not executing query");

$result = @mysql_query($sql, $connection);
echo mysql_error();

aksival

(ignore double post)

tigomark

Thank you all for your help. It was a syntax error with the varible $By. Sql did not like me using one of it's opperators as a varible. I knew it had to be something small. Thank you for your help!!!